MHT CET · Maths · Complex Number
If \(\mathrm{a}\gt0\) and \(\mathrm{z}=\frac{(1+\mathrm{i})^2}{\mathrm{a}-\mathrm{i}}, \mathrm{i}=\sqrt{-1}\), has magnitude \(\sqrt{\frac{2}{5}}\), then \(\overline{\mathrm{z}}\) is equal to
- A \(\frac{1}{5}-\frac{3}{5} \mathrm{i}\)
- B \(-\frac{1}{5}-\frac{3}{5} \mathrm{i}\)
- C \(-\frac{1}{5}+\frac{3}{5} \mathrm{i}\)
- D \(\quad-\frac{3}{5}-\frac{1}{5} \mathrm{i}\)
Answer & Solution
Correct Answer
(B) \(-\frac{1}{5}-\frac{3}{5} \mathrm{i}\)
Step-by-step Solution
Detailed explanation
\(z =\frac{(1+i)^2}{a-i} \)
\( =\frac{2 i}{a-i} \)
\( =\frac{2 i(a+i)}{a^2+1} \)
\( =\frac{-2+2 a i}{a^2+1} \)
\( \therefore |z| =\sqrt{\frac{4+4 a^2}{\left(a^2+1\right)^2}} \)
\( \Rightarrow \sqrt{\frac{2}{5}}=\frac{2}{\sqrt{a^2+1}} \)
\( \Rightarrow a=3\)
\(\therefore z=\frac{-2+2(3) i}{3^2+1}=\frac{-2+6 i}{10} \)
\( \Rightarrow z=\frac{-1+3 i}{5} \)
\( \Rightarrow \bar{z}=\frac{-1}{5}-\frac{3 i}{5}\)
\( =\frac{2 i}{a-i} \)
\( =\frac{2 i(a+i)}{a^2+1} \)
\( =\frac{-2+2 a i}{a^2+1} \)
\( \therefore |z| =\sqrt{\frac{4+4 a^2}{\left(a^2+1\right)^2}} \)
\( \Rightarrow \sqrt{\frac{2}{5}}=\frac{2}{\sqrt{a^2+1}} \)
\( \Rightarrow a=3\)
\(\therefore z=\frac{-2+2(3) i}{3^2+1}=\frac{-2+6 i}{10} \)
\( \Rightarrow z=\frac{-1+3 i}{5} \)
\( \Rightarrow \bar{z}=\frac{-1}{5}-\frac{3 i}{5}\)
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